Carcass wrote:
A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?
(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16
We can solve this question using counting methods.
P(exactly 2 girls and 2 boys) = (number of 4-baby outcomes with exactly 2 girls and 2 boys)/(TOTAL number of 4-baby outcomes) As always, we'll begin with the denominator.
TOTAL number of 4-baby arrangementsThere are
2 ways to have the first baby (boy or girl)
There are
2 ways to have the second baby (boy or girl)
There are
2 ways to have the third baby (boy or girl)
There are
2 ways to have the fourth baby (boy or girl)
By the Fundamental Counting Principle (FCP), the total number of 4-baby arrangements = (
2)(
2)(
2)(
2) =
16Number of 4-baby outcomes with exactly 2 girls and 2 boysThis portion of the question boils down to
"In how many different ways can we arrange 2 G's and 2 B's (where each G represents a girl, and each B represents a boy)?"----------ASIDE-------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....] So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are
11 letters in total
There are
4 identical I's
There are
4 identical S's
There are
2 identical P's
So, the total number of possible arrangements =
11!/[(
4!)(
4!)(
2!)]
------------BACK TO THE QUESTION---------------------------
Our goal is to arrange the letters G, G, B, and B
There are
4 letters in total
There are
2 identical G's
There are
2 identical B's
So, the total number of possible arrangements =
4!/[(
2!)(
2!)] =
6So.....
P(exactly 2 girls and 2 boys) =
6/16 =
3/8Answer: A